设 $\Omega$ 是由球面 $x^2+y^2+z^2=z$ 所围成的闭区域, 求 $\displaystyle\iiint_{\Omega}\sqrt{x^2+y^2+z^2}\mathrm{d}x\mathrm{d}y\mathrm{d}z$.
1
这里的球面方程为 $x^2+y^2+(z-\frac{1}{2})^2=\frac{1}{4}$, 是以 $(0,0,\frac{1}{2})$ 为中心, 半径为 $\frac{1}{2}$ 的球面. 我们采用球面坐标系作变换. 令
\[
\left\{
\begin{aligned}
x&=r\sin\varphi\cos\theta,\\
y&=r\sin\varphi\sin\theta,\\
z&=r\cos\varphi,
\end{aligned}
\right.
\]
其中 $\theta\in[0,2\pi]$, $\varphi\in[0,\frac{\pi}{2}]$, $r\in[0,\cos\varphi\cdot\sin\varphi]$. 于是
\[
\begin{split}
\iiint_{\Omega}\sqrt{x^2+y^2+z^2}\mathrm{d}x\mathrm{d}y\mathrm{d}z&=\int_{0}^{2\pi}\mathrm{d}\theta\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\int_{0}^{\cos\varphi}r\cdot r^2\sin\varphi\mathrm{d}r\\
&=2\pi\int_{0}^{\frac{\pi}{2}}\sin\varphi\mathrm{d}\varphi\cdot(\frac{1}{4}r^4)\biggr|_{0}^{\cos\varphi}\\
&=\frac{\pi}{2}\int_{0}^{\frac{\pi}{2}}\sin\varphi\cdot\cos^4\varphi\mathrm{d}\varphi\\
&=-\frac{\pi}{2}\int_{0}^{\frac{\pi}{2}}\cdot\cos^4\varphi\mathrm{d}\cos\varphi\\
&\stackrel{t=\cos\varphi}{=}-\frac{\pi}{2}\int_1^0 t^4\mathrm{d}t\\
&=\frac{\pi}{2}\int_0^1 t^4\mathrm{d}t\\
&=\frac{\pi}{2}\cdot\frac{1}{5}t^5\biggr|_0^1\\
&=\frac{\pi}{10}.
\end{split}
\]
注意, 如果采用下面的变换, 则比较难以计算.
\[
\left\{
\begin{aligned}
x&=r\sin\varphi\cos\theta,\\
y&=r\sin\varphi\sin\theta,\\
z&=\frac{1}{2}+r\cos\varphi,
\end{aligned}
\right.
\]